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Number of hairstyles you can get from 4 braids, 5 ribbons and 3 barettes? And how do you get the answer? Join or Log In to answer.

-ladyt (2Q / 6A)
Wed Apr 30, 2008 7:03pm

Wouldn't it be 60? 4 x 5 x 3?

-Mangangali (8Q / 6467A) Wed Apr 30, 2008 7:08pm

Does order matter? if so it'll be something like 5! x 4! x 3! (17,280)

-elvis (80Q / 431A) Wed Apr 30, 2008 7:15pm

Comments:

What if all three are required for every hairstyle?

-ladyt (2Q / 6A) Wed Apr 30, 2008 7:17pm

ooh combinatorics, fun,
are we saying that you choose one braid style, one ribbon and one barrette per style? if so 5*4*3=60, can you give the question word for word to be sure?

-nbosch (34Q / 484A) Wed Apr 30, 2008 7:21pm

Comments:

Alicia get her hair braided at a salon. She has a choice of four styles of braid, five ribbon colors and three barrettes. Alicia may choose one style of braid, one ribbon color and one barrette. How many different hairdos can she get?

-ladyt (2Q / 6A) Wed Apr 30, 2008 7:25pm

(To ladyt)
yup, 60 it is do you understand why?

-nbosch (34Q / 484A) Wed Apr 30, 2008 7:30pm

(To nbosch)
http://www.teachers.ash.org.au/mikemath/mathsc/structurecount/index.html
and click on the link for the notes, its a pdf with some good info if you're still stuck... good luck!

-nbosch (34Q / 484A) Wed Apr 30, 2008 7:34pm

(To nbosch)
I understand 4 (braids) x 5 (ribbons) x 3 (barrettes) = 60...but for some reason I think there needs to be another component...like 4x5x3x3=180.

-ladyt (2Q / 6A) Wed Apr 30, 2008 7:40pm

(To ladyt)
yeah thats what got me, if she doesn't have to have all three then the number of styles goes up a bunch

-Mangangali (8Q / 6467A) Wed Apr 30, 2008 7:42pm

(To nbosch)
Many thanks!

-ladyt (2Q / 6A) Wed Apr 30, 2008 7:44pm

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